If a spring is cut in half, does the period change?

Click For Summary

Homework Help Overview

The discussion revolves around the effects of cutting a spring in half on the period of a simple harmonic oscillator, which consists of a mass and an ideal spring characterized by a spring constant. Participants explore how this alteration impacts the oscillation period.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants examine the relationship between the spring constant and the period of oscillation, questioning whether cutting the spring affects the spring constant. Some express confusion about the implications of amplitude changes on the period.

Discussion Status

There is an ongoing exploration of how the spring constant changes when the spring is cut in half, with some participants providing insights into the mathematical relationships involved. A few have reached a clearer understanding of the implications for the period, but no consensus has been explicitly stated.

Contextual Notes

Participants reference previous discussions and archived problems, indicating that there may be varying interpretations of the problem's parameters and assumptions regarding the spring's behavior when altered.

clairez93
Messages
113
Reaction score
0

Homework Statement



A simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. Particle oscillates as shown in (i) with preriod T. If the spring is cut in half and used with the same particle, as shown in (ii), the period will be.

A) 2T
B) \sqrt{2}T
C) T/\sqrt{2}
D) T
E) T/2

Homework Equations



T = 2\pi\sqrt{m/k}

The Attempt at a Solution



I figured that since the length of the spring isn't relevant to the period, the period would stay the same. The answer provided is C. I'm not sure why.
 

Attachments

  • springs.jpg
    springs.jpg
    9.6 KB · Views: 1,681
Physics news on Phys.org
Hm, a thought, does cutting the spring in half alter the constant in some way?
 
This is all I can think of, because the mass stays the same, therefore the only thing that would change is the K, however why would cutting the spring in half change the K?
 
I read this problem in an archive, however, the answers given there don't seem to make sense. They start talking about the amplitude changing, however, isn't the amplitude independent of the period?
 
Consider a spring which extends by length x under a load of mass m.
Connect two such springs in series to form one long spring. What would be the new extension of this spring under a load of mass m? What is the effective new spring constant of this long spring?
Can you extend this idea to cutting a spring in half?
 
The extension would be longer, but how does that affect the spring constant?
 
If you take the full spring and hang a weight W from it, it will stretch an amount Δx. So the spring constant is k = W/Δx.

Now consider the two halves of that spring. How much stretch does each half get? The force (W) is the same, so what must be the spring constant for each half?
 
Oh I see. So the first spring:
k = mg/x

Thus, T = 2\pi\sqrt{x/g}

So in the second spring:

k = 2mg/x

Thus, T' = 2\pi\sqrt{x/2g} = T/\sqrt{2}

Is this correct?
 
Yes, that's right. The key thing is that the spring constant of half a spring is twice the spring constant of the full spring. (This should make a bit of sense. You'd expect it to be harder to pull a shorter spring to the same extension.) And since the period depends on the spring constant, it will change accordingly.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
24
Views
4K
Replies
3
Views
1K
Replies
7
Views
1K
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K